# Thread: Limit of Sequence - max(a,b)

1. ## Limit of Sequence - max(a,b)

Prove that if $\displaystyle a,b$ are positive real numbers then $\displaystyle \displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \max{(a,b)}$.

My attempt is to say that without loss of generality let $\displaystyle a > b$, then for sufficiently large $\displaystyle n$ we will have $\displaystyle a^n + b^n \approx a^n$ hence $\displaystyle \displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \displaystyle\lim_{n\rightarrow \infty}(a^n)^{1/n} = a$. However this step "for sufficiently large $\displaystyle n$, we will have $\displaystyle a^n + b^n \approx a^n$" does not seem rigorous to me, and I can't think of a way of proving this result more rigorously.

Any suggestions would be greatly appreciated.

2. Originally Posted by utopiaNow
Prove that if $\displaystyle a,b$ are positive real numbers then $\displaystyle \displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \max{(a,b)}$.

My attempt is to say that without loss of generality let $\displaystyle a > b$, then for sufficiently large $\displaystyle n$ we will have $\displaystyle a^n + b^n \approx a^n$ hence $\displaystyle \displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \displaystyle\lim_{n\rightarrow \infty}(a^n)^{1/n} = a$. However this step "for sufficiently large $\displaystyle n$, we will have $\displaystyle a^n + b^n \approx a^n$" does not seem rigorous to me, and I can't think of a way of proving this result more rigorously.

Any suggestions would be greatly appreciated.

As you did, assume wrg that a>b and then "take a out of the parentheses" :

$\displaystyle \sqrt[n]{a^n+b^n}=a\sqrt[n]{1+\left(\frac{b}{a}\right)^n}$ and remember what happens when we have a geometric sequence with quotient q s.t. $\displaystyle |q|<1$

Tonio